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memicm105
@memicm105
January 2021
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Pouvez vous m'aider pour ce dm s'il vous plaît
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taalbabachir
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1) calculer sin (x)
sin (x) = JH/SH
Il faut chercher JH et SH
SA/SJ = AF/JH ⇔ SA x JH = SJ x AF ⇒ JH = SJ x AF/SA = 12 x 2/6 = 4 m
Appliquons le théorème de Pythagore
SH² = SJ² + JH² = 12² + 4² = 144 + 16 = 160 ⇒ SH = √160 = 4√10 m
sin (x) = JH/SH = 4/4√10 = 1/√10
2) en déduire les 2 valeurs possibles de cos (x)
cos² (x) + sin² (x) = 1 ⇒ cos² (x) = 1 - sin²(x) = 1 - (1/√10)² = 1 - 1/10 = 9/10
cos² (x) = 9/10 ⇒ cos (x) = √9/10 et cos (x) = - √9/10
cos (x) = 3/√10 ; cos (x) = - 3/√10
3) calculer la tan (x) en fonction de la longueur HJ
tan (x) = HJ/SH = HJ/12
4) en déduire HJ en fonction de tan (x)
HJ/12 = tan (x) ⇒ HJ = 12 x tan (x)
5) en déduire les 2 valeurs possibles de tan (x) et donc les deux valeurs de HJ possibles
tan (x) = sin (x)/cos (x)
sin (x) = 1/√10
cos (x) = cos (x) = 3/√10 ; cos (x) = - 3/√10
tan (x) = 1/√10/3/√10 = 1/3
tan (x) = -1/√10/3/√10 = -1/3
HJ = 12 x tan (x) = 12 x 1/3 = 4 m
HJ = 12 x tan (x) = - 12 x 1/3 = - 4 m
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Verified answer
1) calculer sin (x)sin (x) = JH/SH
Il faut chercher JH et SH
SA/SJ = AF/JH ⇔ SA x JH = SJ x AF ⇒ JH = SJ x AF/SA = 12 x 2/6 = 4 m
Appliquons le théorème de Pythagore
SH² = SJ² + JH² = 12² + 4² = 144 + 16 = 160 ⇒ SH = √160 = 4√10 m
sin (x) = JH/SH = 4/4√10 = 1/√10
2) en déduire les 2 valeurs possibles de cos (x)
cos² (x) + sin² (x) = 1 ⇒ cos² (x) = 1 - sin²(x) = 1 - (1/√10)² = 1 - 1/10 = 9/10
cos² (x) = 9/10 ⇒ cos (x) = √9/10 et cos (x) = - √9/10
cos (x) = 3/√10 ; cos (x) = - 3/√10
3) calculer la tan (x) en fonction de la longueur HJ
tan (x) = HJ/SH = HJ/12
4) en déduire HJ en fonction de tan (x)
HJ/12 = tan (x) ⇒ HJ = 12 x tan (x)
5) en déduire les 2 valeurs possibles de tan (x) et donc les deux valeurs de HJ possibles
tan (x) = sin (x)/cos (x)
sin (x) = 1/√10
cos (x) = cos (x) = 3/√10 ; cos (x) = - 3/√10
tan (x) = 1/√10/3/√10 = 1/3
tan (x) = -1/√10/3/√10 = -1/3
HJ = 12 x tan (x) = 12 x 1/3 = 4 m
HJ = 12 x tan (x) = - 12 x 1/3 = - 4 m